Question 1193138
<pre>
There is no way to use methods of algebra to solve an equation for 
an unknown that appears both inside and outside of a logarithmic, 
exponential, or trigonometric function.  The only way is to 
approximate it with a computer, calculator, or iterative method 
("educated" trial and error).

Perhaps you did not mean to have " x= " in the front as if it were
part of the equation, and meant only this equation:

{{{log(2,(x))-log(8,(x))}}}{{{""=""}}}{{{4}}}

Let's convert one of the logs so we will have both logs to the same 
base.

I choose to convert {{{log(8,(x))}}} to base 2.

You should memorize the change of base formula:

 {{{log(b,(a))=log(c,(a))/log(c,(b))}}}

and plug in a=x, b=8, c=2

{{{log(8,(x))=log(2,(x))/log(2,8)}}}{{{""=""}}}{{{log(2,(x))/log(2,2^3)=log(2,x)/3}}}

So substitute {{{log(2,(x))/3}}} for {{{log(8,(x))}}} in

{{{log(2,(x))-log(8,(x))}}}{{{""=""}}}{{{4}}}


{{{log(2,(x))-log(2,(x))/3^""^""}}}{{{""=""}}}{{{4}}}


Multiply through by 3

{{{3log(2,(x))-log(2,(x))}}}{{{""=""}}}{{{12}}}

{{{2log(2,(x))}}}{{{""=""}}}{{{12}}}

{{{log(2,(x))}}}{{{""=""}}}{{{6}}}

By the definition of logarithms, from the above equation,
we have this:

{{{x}}}{{{""=""}}}{{{"_____"}}}{{{""=""}}}{{{"_____"}}} 

(You fill in those blanks, the first one with an 
eponential expression and the second with a number.

Edwin</pre>